YT14-HDU-找循环节 （关于std::ios::sync_with_stdio(false);的作用和疑问）

Problem Description

As a unicorn, the ability of using magic is the distinguishing feature among other kind of pony. Being familiar with composition and decomposition is the fundamental course for a young unicorn. Twilight Sparkle is interested in the decomposition of permutations. A permutation of a set S = {1, 2, ..., n} is a bijection from S to itself. In the great magician —— Cauchy's two-line notation, one lists the elements of set S in the first row, and then for each element, writes its image under the permutation below it in the second row. For instance, a permutation of set {1, 2, 3, 4, 5} σ can be written as:


Here σ(1) = 2, σ(2) = 5, σ(3) = 4, σ(4) = 3, and σ(5) = 1.
Twilight Sparkle is going to decompose the permutation into some disjoint cycles. For instance, the above permutation can be rewritten as:


Help Twilight Sparkle find the  lexicographic smallest solution. (Only considering numbers).

Input

Input contains multiple test cases (less than 10). For each test case, the first line contains one number n (1<=n<=10^5). The second line contains n numbers which the i-th of them(start from 1) is σ(i).

Output

For each case, output the corresponding result.

Sample Input

5
2 5 4 3 1
3
1 2 3

Sample Output

(1 2 5)(3 4)
(1)(2)(3)

原代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
const int NUM=100000;
using namespace std;
int main()
{
    int i,j,n,m;
    int a[NUM];
    while(cin>>n)
    {
        for(i=1; i<=n; i++)
            cin>>a[i];
        for(i=1; i<=n; i++)
        {
            while(a[i])
            {
                cout<<"("<<i;
                j=a[i];
                a[i]=0;
                while(a[j])
                {
                    cout<<" "<<j;
                    m=a[j];
                    a[j]=0;
                    j=m;
                }
                cout<<")";
            }
        }
        cout<<endl;
    }
    return 0;
}

AC代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
const int NUM=100000;
using namespace std;
int main()
{
    std::ios::sync_with_stdio(false);
    int i,j,n,m;
    int a[NUM];
    while(cin>>n)
    {
        for(i=1; i<=n; i++)
            cin>>a[i];
        for(i=1; i<=n; i++)
        {
            while(a[i])
            {
                cout<<"("<<i;
                j=a[i];
                a[i]=0;
                while(a[j])
                {
                    cout<<" "<<j;
                    m=a[j];
                    a[j]=0;
                    j=m;
                }
                cout<<")";
            }
        }
        cout<<endl;
    }
    return 0;
}

运行结果：

提交自己的代码之后经常是Time Limit Exceeded。

看网上找代码的时候别人用的通常都是C语言的scanf和printf输入输出，一直以为那是C语言的代码，但这次偶然看到了同学的代码（和我“抄”得居然是同一个。。。老师教导我们要”抄“之有道）但其用了

于是百度了一下：

cin，cout之所以效率低，是因为先把要输出的东西存入缓冲区，再输出，导致效率降低，而这段语句可以来打消iostream的输入输出缓存，可以节省许多时间，使效率与scanf与printf相差无几。

又学到了一手，最近做题总算出现时间超限，以后就不用担心了。不过不知道

std::ios::sync_with_stdio(false);

有什么弊端，